Find the smallest number which when increased by 20 , is exactly divisible by 72, 90 and 150 .

CBSE Class 10 Maths PYQ · Real Numbers · Word Problem · 3 Marks · March 2025 · Basic

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1633 Marks · March 2025 · Basic
Find the smallest number which when increased by $20$, is exactly divisible by $72, 90$ and $150$.
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Solution: $72 = 2^3 \times 3^2, 90 = 3^2 \times 2 \times 5, 150 = 5^2 \times 2 \times 3$
$LCM (72, 90, 150) = 2^3 \times 3^2 \times 5^2 = 1800$
Required smallest number is $1800 - 20 = 1780$
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